Question

Solve the equation `9x^2 + (3x)/4 + 2 = 0`, if possible, for real values of x.

Answer 1

`9x^2 + (3x)/4 + 2 = 0`

`=> (36x^2 + 3x + 8)/4 = 0`

`=> 36x^2 + 3x + 8 = 0`

Here a = 36, b = 3 and c = 8

`∴ x = (-b +- sqrt(b^2 - 4ac))/(2a)`

`= (-3 +- sqrt((3)^2 - 4 xx 36 xx 8))/(2 xx 36)`

`= (-3 +- sqrt(9 - 1152))/72`

`= (-3 +- sqrt(-1143))/72`

Since `sqrt(-1143)` is not possible, we cannot solve the given equation for x.